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This is an exercise from Kunen - An introduction to independence proofs that I have hard time to solve.

In $ZF$, $AC$ is equivalent to $\forall \alpha (\mathscr P(\alpha)$ can be well-ordered)

Any advice would be welcomed!

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marked as duplicate by Asaf Karagila axiom-of-choice Dec 9 '15 at 19:23

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  • $\begingroup$ Pretty sure this is a duplicate. Will look for it soon. $\endgroup$ – Asaf Karagila Dec 9 '15 at 19:17
  • $\begingroup$ This is not the same question, your answer claimed that "$\forall X$ if $X$ is well-orderable then so is $2^X$" implies AC. $\endgroup$ – nombre Dec 9 '15 at 22:09
  • $\begingroup$ @mathcounterexamples.net: AC implies the other result since it implies Zermelo's theorem which is $\forall \alpha$, $\alpha$ can be well-ordered. As for the other implication, if there is no restriction on $\alpha$ (if ou don't want it to be an ordinal) you can start by proving that if $\alpha$ embeds in a well-orderable set then it is well-orderable, then use Zermelo $\longrightarrow$ AC. In any case, the difficult result is the equivalency of Zermelo's theorem and AC. $\endgroup$ – nombre Dec 9 '15 at 22:16